Prove that a closed subset E of a metric space (M,d) is disconnected iff there exists disjoint nonempty closed sets E1, E2 such that E=E1UE2.
for the --> direction, we have E is disconnected, then by definition there exists nonempty disjoint open sets A,B s.t E=AUB, how do I turn this into a union of 2 closed sets, we can't just say since E is closed, then its subsets is closed, right?
also need some hint for the other direction.
ok, I have done one direction, I need to do the --> direction now, supposing E is disconnected, there exists disjoint E1,E2 subset of E, such that E=E1UE2, and there exist open sets A,B such that E1 subset of A and E2 subset of B.
We need to show that E1, E2 are closed, since we know E is closed, what can we say about E1 and E2? E=E1UE2 implies E1, E2 are closed?
sorry for the confusion, now i need to prove if a closed subset E of a metric space (M,d) is disconnected , then there exists disjoint nonempty closed sets E1, E2 such that E=E1UE2.
so supposing E is disconnected, there exists disjoint E1,E2 subset of E, such that E=E1UE2, and there exist open sets A,B such that E1 subset of A and E2 subset of B.
We need to show that E1, E2 are closed, since we know E is closed, what can we say about E1 and E2? E=E1UE2 implies E1, E2 are closed?
I think I know what definition of connect set you are using.
If is not connected the there exists two disjoint open sets each having non-empty intersection with such that .
Let . Clearly .
Now if is a limit point of then is a limit poiint of .
Therefore, because of disjoint open sets.